In the previous section, we saw some solved examples on section formula. In this section, we will see some miscellaneous examples.
Solved example 12.11
Show that the points A(1,2,3), B(-1,-2,-1), C(2,3,2) and D(4,7,6) are the vertices of a parallelogram ABCD but it is not a rectangle.
Solution:
1. A rough sketch is shown in fig.12.19 below:
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| Fig.12.19 |
2. Using the distance formula, we can find the lengths of sides.
♦ AB = √36 = 6
♦ BC = √43
♦ CD = √36 = 6
♦ DA = √43
3. We see that, both pairs of opposite sides (AB, CD) and (BC, DA) are equal. So it is a parallelogram.
4. A rectangle is a parallelogram in which both diagonals are of the same length.
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If a parallelogram is not to be a rectangle, it's diagonals should not be of the same length.
5. Using distance formula, let us write the lengths of the diagonals:
♦ AC = √3
♦ BD = √155
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We see that, diagonals are not equal. So it a parallelogram which is not a rectangle.
Solved example 12.12
Find the equation of the set of points P such that it's distances from the points A(3,4,-5) and B(-2,1,4) are equal.
Solution:
1. Let P(x,y,z) be equidistant from A and B
2. Let us use the distance formula.
• Square of the distance PA can be obtained as follows:
PA2 = (x-3)2 + (y-4)2 + (z+5)2
= x2 -6x +9 +y2 -8y +16 +z2 +10z +25
• Square of the distance PB can be obtained as follows:
PB2 = (x+2)2 + (y-1)2 + (z-4)2
= x2 +4x +4 +y2 -2y +1 +z2 -8z +16
3. Since the distances are equal, the square of the distances will also be equal. We can write:
x2 -6x +9 +y2 -8y +16 +z2 +10z +25 = x2 +4x +4 +y2 -2y +1 +z2 -8z +16
⇒ -6x +9 -8y +16 +10z +25 = +4x +4 -2y +1 -8z +16
⇒ -6x -4x +9 -8y +2y +16 +10z +8z +25 = +4 +1 +16
⇒ -10x +9 -6y +16 +18z +25 = +4 +1 +16
⇒ -10x -6y +18z +50 = 21
⇒ -10x -6y +18z +29 = 0
⇒ 10x +6y -18z -29 = 0
Solved example 12.13
The centroid of a triangle ABC is at G(1,1,1). If the coordinates of A and B are (3,-5,7) and (-1,7,-6) respectively, find the coordinates of C.
Solution:
1. Let the coordinates of C be (x,y,z)
Then we can draw a rough sketch as shown below:
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| Fig.12.20 |
2. We know that, the coordinates of the centroid of any triangle can be obtained using the expression:
$\left( \frac{x_1 + x_2 + x_3}{3},~ \frac{y_1 + y_2 + y_3}{3},~ \frac{z_1 + z_2 + z_3}{3}\right)$
3. Substituting the known values, we get:
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$ \frac{3 -1 + x}{3}~=~1$
⇒ 2+x = 3
⇒ x = 1
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$ \frac{-5 +7 + y}{3}~=~1$
⇒ 2+y = 3
⇒ y = 1
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$ \frac{7 -6 + z}{3}~=~1$
⇒ 1+x = 3
⇒ z = 2
4. So the coordinates of C are: (1,1,2)
The link below gives a few more solved examples:
We have completed a disscusion on three dimensional geometry. In the next chapter, we will see limits and derivatives.
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